Question: Chapter 2. Show that the matrix $A$ satisfies Cayley-Hamilton Theorem and hence find $A^{-1}$, where
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$A=\begin{bmatrix} 1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1 \end{bmatrix}$

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Solution:

$A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$

$|A|=1$

Characteristic equation is $|A-\lambda I|=0$

$\begin{vmatrix} 1-\lambda & 2 & -2 \\ -1 & 3-\lambda & 0 \\ 0 & -2 & 1-\lambda \end{vmatrix}=0$

$\lambda^{3}-(5) \lambda^{2}+(3+3+1+2-0-0)\lambda-1=0$

$\lambda^{3}-5\lambda^{2}+9\lambda-1=0$ -------------(1)

Cayley-Hamilton Theorem states that this equation is satisfied by the matrix A.

$\therefore$ Equation (1) becomes $A^{3}-5A^{2}+9A-I=0$ -------------(2)

$A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ ---------------(3)

$A^{2}=A.A=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}$ -------------(4)

$A^{3}=A.A^{2}=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}=\begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}$ -------------(5)

$I=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ ---------------(6)

Put (3), (4), (5), (6) in (2),

$\begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}-5\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}+9\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\therefore$ Cayley-Hamilton Theorem verified.

To find $A^{-1} \Longrightarrow$

Multiply equation (2) by $A^{-1}$

$(A^{3})A^{-1}-5A^{2}(A^{-1})+9A(A^{-1})-I(A^{-1})=0$

$A^{2}-5A+9I-A^{-1}=0$ ----------------(7)

Put equation (3), (4), (6) in (7),

$A^{-1}=\begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} -5 \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}+9\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$